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ra1l [238]
4 years ago
8

Which function forms an arithmetic sequence when x = 1, 2, 3, ...?

Mathematics
1 answer:
____ [38]4 years ago
5 0
When we have power function or exponential, or then we have a/x, then functions are not going to give arithmetic sequence, because an arithmetic sequence it is really a linear function  when x are natural numbers,

so we need to check  only A
<span>f(x)=- x/3 +5
 
x  1             2               3              4              5
y 14/3      13/3         12/3        11/3        10/3           arithmetic difference is -1/3

so the answer is A</span>
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The present age of Andy and Mike are in the ratio 5:3. If Andy had been 7 years older and Mike 7 years younger, the age of Andy
Tju [1.3M]

Answer: Andy is 35 and Mike is 21.

Step-by-step explanation:

Andy’s present age: Mike’s present age

= 5:3

Andy’s age in “if”: Mike’s age in “if”

=6:2

1 unit = 7

5 unit = 35 (Andy’s age)

3 unit = 21 (Mike’s age)

6 0
3 years ago
Louisa has $145 saved in her account. She makes $8 an hour at work. How many hours must she work to have at least $320?
musickatia [10]
<span>$145 saved in her account.
</span><span>$8 an hour at work.
</span><span>at least $320

320 - 145 = 175

175 </span>÷ 8 = 21.875

Louisa needs to work about 22 more hrs to have at least $320!
6 0
3 years ago
1.2 Puzzle Time
ruslelena [56]
B because it is the answer
4 0
3 years ago
Read 2 more answers
A fishing supply store buys fishing rods for $11 and sells them at a markup of 220% One day, they decide to put the rods on sale
Alina [70]

Answer:28.16

Step-by-step explanation:

A markup of 220% would bring 11 to 35.20, and a mark down of 20% would bring 35.20 to 28.16.

Your'e Welcome  (ФДФ) .

8 0
3 years ago
Which row operation will triangularize this matrix?
Ludmilka [50]
Triangularizing matrix gives the matrix that has only zeroes above or below the main diagonal. To find which option is correct we need to calculate all of them.
In all these options we calculate result and write it into row that is first mentioned:

A)R1-R3
\left[\begin{array}{ccc}-1&0&0|0\\0&1&1|6\\2&0&1|1\end{array}\right]

B)2R2-R3
\left[\begin{array}{ccc}1&0&1|1\\-2&2&1|4\\2&0&1|1\end{array}\right]

C)-2R1+R3
\left[\begin{array}{ccc}0&0&-1|-1\\0&1&1|6\\2&0&1|1\end{array}\right]

D)2R1+R3
\left[\begin{array}{ccc}4&0&3|3\\0&1&1|6\\2&0&1|1\end{array}\right]

E)3R1+R3
\left[\begin{array}{ccc}5&0&4|4\\0&1&1|6\\2&0&1|1\end{array}\right]

None of the options will triangularize this matrix. The only way to <span>triangularize this matrix is
R3-2R1
</span>\left[\begin{array}{ccc}1&0&1|1\\0&1&1|6\\0&0&-1|-1\end{array}\right]
<span>
This equation is similar to C) but in reverse order. Order in which rows are written is important.</span>
4 0
3 years ago
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